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Notamathematician
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comment
Elegant recursion for A301897
@MartinRubey, try $b(2^m(2k+1))=-b(2^{m+1}k)+\sum\limits_{j=0}^{m+m\operatorname{mod}3+1}b(2^j k)$ for $m\geqslant 0, k\geqslant 0$ with $b(0)=1$.
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Something (which might be called multi-continued fraction) for the A112487
@AlexanderBurstein, sorry, I calculated incorrectly. You are absolutely right.
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Something (which might be called multi-continued fraction) for the A112487
@PeterTaylor, thank you for comment! See PARI prog. By the way, if you try to reduce $G(j+1)$, it does not give a result.
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Something (which might be called multi-continued fraction) for the A112487
@MaxAlekseyev, thank you for comment! $G(0)$ is completely defined by a given recurrence. Just put $j=0$ and then expand $G(1)$ using $G(2)$, then expand $G(2)$ using $G(3)$ and so on.
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Recursion for the A266328 by analogy with non-standard recursion for factorials
@LSpice, thank you for comment and for editing too! I don't know much about integrals, so I took the definition of $A(x)$ from A266328.
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